In this experiment, you will investigate the equilibrium position of a suspended cardboard sheet. (a) You are provided with a flat sheet of cardboard with a hole through it and two lines drawn near two of the edges. Assemble the apparatus as shown in Fig. 1.1 with the bottom edge of the cardboard approximately $5\text{ cm}$ above the bench. Check that the cardboard swings freely on the knitting needle. Use the sharp pencil to make a hole through the cardboard approximately…

(a) Example: x = 8.0 cm, θ = 19° · (b) See working / student-dependent (table of x, θ and 1/tanθ with 6 sets) · (c)(i) Graph plotted: 1/tanθ (y) against x (x) · (c)(ii) Straight line of best fit drawn · (c)(iii) Example: gradient ≈ 0.30 cm⁻¹, y-intercept ≈ 0.44 · (d) Example: a ≈ 0.30 cm⁻¹, b ≈ 0.44

In this experiment, you will investigate the motion of a conical pendulum. (a) Set the compasses to a radius of approximately $9\text{ cm}$ and then use them to draw a circle on the sheet of paper. Mark the centre of the circle with a cross. Measure and record the diameter $D$ of the circle. $D$ = \_\_\_\_\_\_ $\text{cm}$ (i) You are provided with a pendulum bob with a length of string attached. Tie a knot in the string approximately $19\text{ cm}$ from the top of the bob.…

(a) D = 18.0 cm (example) · (b)(i) p = 20.0 cm (example) · (b)(ii) 0.5% · (c)(i) T = 0.850 s (example) · (c)(ii) 26.7° · (d) Example: p = 14.0 cm, T = 0.660 s, \Phi = 40.1° · (e)(i) k1 = 4.04 s^2 m^-1, k2 = 4.07 s^2 m^-1 (example) · (e)(ii) 2 s.f. (because uncertainty ~15%) · (f) Supports the relationship (agreement within 15%). · (g)(i) See working (four uncertainties/limitations listed). · (g)(ii) See working (four improvements listed).

Physics 9702/36 — October/November 2024

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the equilibrium position of a suspended cardboard sheet.

(a)

You are provided with a flat sheet of cardboard with a hole through it and two lines drawn near two of the edges.

Assemble the apparatus as shown in Fig. 1.1 with the bottom edge of the cardboard approximately $5\text{ cm}$ above the bench. Check that the cardboard swings freely on the knitting needle.

Fig. 1.1

Use the sharp pencil to make a hole through the cardboard approximately half-way along the longer line.
Pass the bolt through the slotted mass and then through the hole in the cardboard, as shown in Fig. 1.2.
Secure the bolt using the nut.

Fig. 1.2

The distance between the centre of the slotted mass and the intersection of the two lines is $x$ , as shown in Fig. 1.2.

Measure and record $x$ .

$x$ = ______

The angle between the bottom edge of the cardboard and the horizontal is $\theta$ , as shown in Fig. 1.2.

Use the wooden block and the protractor to measure $\theta$ .

$\theta$ = ______ $^\circ$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

(Example set of readings)

x = 8.0\ \text{cm}

\theta = 19^\circ

Final answer

Example: x = 8.0 cm, θ = 19°

Detailed explanation

Background Concept

In this experiment you are measuring two quantities:

A distance $x$ (a length), which should be recorded with a sensible precision for a ruler (typically to the nearest $1\ \text{mm}$ , i.e. $0.1\ \text{cm}$ ).
An angle $\theta$ , measured with a protractor relative to a horizontal reference line. Protractors usually allow readings to the nearest $1^\circ$ .

Good practical marks often depend on clear method (how you aligned the instrument) and appropriate recording (units and precision).

Understanding the Question

You are told that:

$x$ is the distance between the centre of the slotted mass and the intersection of the two drawn lines on the cardboard.
$\theta$ is the angle between the bottom edge of the cardboard and the horizontal.

You must measure and record one value of $x$ and one value of $\theta$ for the initial hole position.

Approach

Identify the two reference points for $x$ and use a ruler to measure the straight-line distance between them.
Create a horizontal reference (using the wooden block/bench edge) and use the protractor to measure the angle between the bottom edge of the cardboard and the horizontal.
Record with units and appropriate precision.

Step-by-Step Reasoning

For $x$ :
- Locate the intersection point of the two drawn lines.
- Locate the centre of the slotted mass (estimate its centre as the midpoint).
- Place the ruler so that its scale lies along the line joining these two points.
- Read $x$ at eye level to reduce parallax.
- Record as, for example, $x = 8.0\ \text{cm}$ (to $0.1\ \text{cm}$ ).
For $\theta$ :
- Use the wooden block to provide a clear horizontal reference line.
- Place the protractor with its baseline aligned to the horizontal.
- Read the angle to the nearest degree where the bottom edge of the cardboard intersects the protractor scale.
- Record as, for example, $\theta = 19^\circ$ .

Key Takeaways

Lengths: clear definition of endpoints, avoid parallax, sensible precision.
Angles: need a reliable horizontal reference and correct alignment of the protractor.

Common Mistakes

Measuring $x$ from the edge of the slotted mass instead of its centre.
Forgetting units (e.g. writing $8.0$ instead of $8.0\ \text{cm}$ ).
Reading the wrong scale on the protractor (inner vs outer scale).
Not using a horizontal reference, so $\theta$ is not measured from the horizontal.

Things to Be Careful About

Keep the cardboard swinging freely and let it come to rest before measuring $\theta$ .
Align your eye directly over the ruler/protractor markings to avoid parallax.
Record $x$ and $\theta$ to consistent, appropriate precision (e.g. $0.1\ \text{cm}$ and $1^\circ$ ).

Techniques used

measure a length using a ruler while avoiding parallaxmeasure an angle using a protractor with a horizontal referencerecord readings to an appropriate precision with units

(b)

Use the pencil to make another hole through the longer line and move the slotted mass and bolt to the new hole. Measure $x$ and $\theta$ .

Repeat until you have six sets of values of $x$ and $\theta$ .

Record your results in a table. Include values of $\frac{1}{\tan \theta}$ in your table.

10M Similar

DifficultyMedium

Worked solution

Answer

Record six pairs of $x$ and $\theta$ and calculate $1/\tan\theta$ for each.

(Example of an acceptable table format and typical values)

$x / \text{cm}$	$\theta / ^\circ$	$\dfrac{1}{\tan\theta}$
4.0	31	1.66
6.0	24	2.25
8.0	19	2.90
10.0	16	3.49
12.0	14	4.01
14.0	12	4.70

Final answer

See working / student-dependent (table of x, θ and 1/tanθ with 6 sets)

Detailed explanation

Background Concept

In Paper 3, marks for a results table are typically awarded for:

Having enough readings (here: six sets of $x$ and $\theta$ ).
A sensible range of the independent variable (here $x$ should change noticeably each time).
A single clear table with correct headings (quantity and unit) and consistent decimal places.
Correct calculation of a derived quantity (here $1/\tan\theta$ ).

Note that $\tan\theta$ is dimensionless, so $1/\tan\theta$ is also dimensionless.

Understanding the Question

You must:

Make new holes along the longer line to change the position of the slotted mass.
For each position, measure and record $x$ and $\theta$ .
Produce a table that also contains $1/\tan\theta$ .

So you need three columns: $x$ , $\theta$ , and $1/\tan\theta$ .

Approach

Treat $x$ as the quantity you vary (independent variable).
For each $x$ , measure $\theta$ once the cardboard has come to rest.
Use a calculator to compute $1/\tan\theta$ .
Present everything in one table with correct headings and consistent precision.

Step-by-Step Reasoning

Choose six hole positions spread out along the longer line so that $x$ values are well separated.
For each hole position:
- Allow the cardboard to settle at equilibrium.
- Measure $x$ to (typically) $0.1\ \text{cm}$ .
- Measure $\theta$ to the nearest $1^\circ$ .
Calculate the derived quantity:

\frac{1}{\tan\theta}

Enter $\theta$ in degrees (check calculator mode).
Calculate $\tan\theta$ , then take the reciprocal.
Record $1/\tan\theta$ to a consistent number of significant figures (often 3 s.f. is suitable).

Table conventions:
- Put units in the headings (e.g. $x/\text{cm}$ , $\theta/^\circ$ ).
- Keep decimal places consistent down a column (e.g. all $x$ to 0.1 cm).

Key Takeaways

A good practical table is about clarity: headings with units, consistent precision, and enough readings.
Derived quantities must be calculated correctly and recorded consistently.

Common Mistakes

Omitting units in headings (e.g. writing just $x$ instead of $x/\text{cm}$ ).
Mixing precision (e.g. $x = 4\ \text{cm}$ then $x = 6.3\ \text{cm}$ in the same column).
Forgetting to include the derived column $1/\tan\theta$ .
Calculator in radians instead of degrees, giving wrong $\tan\theta$ values.

Things to Be Careful About

Use a wide enough spread of $x$ to make a clear graph later.
Ensure the cardboard is not touching anything and is truly at rest before reading $\theta$ .
Since $\tan\theta$ changes rapidly at some angles, small reading errors in $\theta$ can noticeably affect $1/\tan\theta$ —be as consistent as possible when aligning the protractor.

Techniques used

collect a suitable range of values by varying the independent variablerecord results in a single table with headings including unitscalculate derived quantities using a calculator and consistent significant figuresrepeat measurements systematically for multiple trials

(c)

(i)

Plot a graph of $\frac{1}{\tan \theta}$ on the $y$ -axis against $x$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $y = \dfrac{1}{\tan\theta}$ against $x$ .

$x$ -axis: $x / \text{cm}$
$y$ -axis: $\dfrac{1}{\tan\theta}$ (no unit)
Use a suitable scale and plot all six points accurately.

Final answer

Graph plotted: 1/tanθ (y) against x (x)

Detailed explanation

Background Concept

A graph is used to show how one quantity depends on another. Good graph marks usually require:

Correct choice of axes (which variable on which axis).
Axes labelled with quantity and unit (if unit exists).
A scale that uses at least half the grid in each direction.
Accurate plotting with fine points (small crosses) and no thick blobs.

Here, $\dfrac{1}{\tan\theta}$ is dimensionless, so the $y$ -axis has no unit.

Understanding the Question

You are told explicitly to plot:

$\dfrac{1}{\tan\theta}$ on the $y$ -axis
$x$ on the $x$ -axis

using the data you collected in part (b).

Approach

Decide the minimum and maximum values of $x$ and $1/\tan\theta$ from your table.
Choose simple scales (e.g. 1 big square = 1 cm, or 1 big square = 0.5) so the points spread out.
Label axes correctly and plot each pair $(x,\ 1/\tan\theta)$ .

Step-by-Step Reasoning

Draw axes with a sharp pencil and mark a clear origin (it does not have to be (0,0) if your data do not include values near zero, but the scale must be clearly shown).
Label:
- Horizontal: $x/\text{cm}$
- Vertical: $1/\tan\theta$
Mark scales at regular intervals.
For each row in the table:
- Find the $x$ value on the horizontal axis.
- Find the corresponding $1/\tan\theta$ value on the vertical axis.
- Plot a small cross at the intersection.

Key Takeaways

Correct axis choice and labelling are essential.
Good use of graph paper and accurate plotting are what earn marks.

Common Mistakes

Swapping axes (plotting $x$ on $y$ -axis).
Missing units in $x$ axis label.
Using an awkward scale (e.g. 1 big square = 3 cm) making plotting inaccurate.
Plotting dots too large, hiding the true position.

Things to Be Careful About

$\theta$ must be in degrees when you calculated $1/\tan\theta$ ; otherwise the plotted values will be wrong.
Do not join the points dot-to-dot; you will draw a best-fit line in the next part.

Techniques used

choose appropriate axis scales that use most of the graph gridlabel axes with quantity and unitplot measured points accurately with small, neat crosses

(ii)

Draw the straight line of best fit.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Draw a single straight line of best fit with approximately equal scatter of points on either side of the line.

Final answer

Straight line of best fit drawn

Detailed explanation

Background Concept

A line of best fit is drawn to represent the trend in experimental data when you expect an approximately linear relationship. It should:

Be a single straight line.
Pass through the middle of the scatter (roughly equal points above and below).
Not be forced through every point.

Understanding the Question

You have plotted $y = 1/\tan\theta$ against $x$ . You must now draw the best straight line representing the trend.

Approach

Use a ruler.
Position it so the line is as balanced as possible with respect to the plotted crosses.
Draw the line across the full range of the data.

Step-by-Step Reasoning

Visually assess the cluster of points.
Place the ruler so that the line goes through the centre of the point pattern.
Ensure that the line is not chosen simply to pass through the first and last points unless that also balances the other points.
Draw a thin, continuous line spanning the plotted range.

If one point is clearly far away from the trend (an anomaly), the best-fit line should still follow the majority of points.

Key Takeaways

Best fit means “most representative”, not “connect the dots”.
The line should be long enough for accurate gradient calculation later.

Common Mistakes

Drawing dot-to-dot segments instead of one straight line.
Forcing the line through the origin without evidence.
Choosing a line that leaves almost all points on one side.

Things to Be Careful About

Keep the line thin and neat (thick lines reduce accuracy when reading values for gradients/intercepts).
Extend the line across most of the graph so you can use widely spaced points to find the gradient accurately.

Techniques used

judge the overall linear trend of plotted datadraw a balanced straight line of best fit through the scatterignore anomalous points when justified by scatter pattern

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______

$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Choose two well-separated points on the best-fit line, e.g.

(x_1, y_1) = (4.0\ \text{cm}, 1.66),\quad (x_2, y_2) = (14.0\ \text{cm}, 4.70)

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{4.70 - 1.66}{14.0 - 4.0} = 0.304\ \text{cm}^{-1}

y\text{-intercept} = y - (\text{gradient})x = 1.66 - (0.304)(4.0) = 0.44

Answer

\text{gradient} \approx 0.30\ \text{cm}^{-1}

y\text{-intercept} \approx 0.44

Final answer

Example: gradient ≈ 0.30 cm⁻¹, y-intercept ≈ 0.44

Detailed explanation

Background Concept

For a straight-line graph, the key quantities are:

Gradient (slope):

\text{gradient} = \frac{\Delta y}{\Delta x}

$y$ -intercept: the value of $y$ when $x=0$ (where the line crosses the $y$ -axis).

To reduce uncertainty, you should use a large triangle (points far apart on the line), and you should use points on the line of best fit, not necessarily the original plotted points.

Understanding the Question

You must use your graph of $1/\tan\theta$ (vertical) against $x$ (horizontal) to find:

the gradient of the best-fit line
the $y$ -intercept of the best-fit line

These values are then used in part (d).

Approach

Pick two points that lie exactly on the best-fit line and are widely separated.
Read their coordinates $(x_1,y_1)$ and $(x_2,y_2)$ from the graph.
Compute the gradient using $\Delta y/\Delta x$ .
Find the intercept either by reading it directly at $x=0$ or by substituting one point into $y = mx + c$ .

Step-by-Step Reasoning

Suppose you choose two points on the line:

(x_1, y_1),\ (x_2, y_2)

Calculate changes:

\Delta y = y_2 - y_1,\quad \Delta x = x_2 - x_1

Then:

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}

Units: since $y = 1/\tan\theta$ is dimensionless and $x$ is in $\text{cm}$ , gradient has units $\text{cm}^{-1}$ .

To find the intercept $c$ :

y = mx + c \implies c = y - mx

where $m$ is the gradient.

Alternatively, you can extend the line back to $x=0$ and read the intercept directly (but calculation often gives a cleaner value).

Key Takeaways

Use points on the best-fit line, not raw data points.
Use a large triangle for a more accurate gradient.
Remember units: gradient carries units; intercept carries the same units as $y$ .

Common Mistakes

Using $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Choosing points that are too close together, giving a large percentage uncertainty.
Reading coordinates from plotted points rather than from the best-fit line.
Forgetting the gradient unit ( $\text{cm}^{-1}$ ).

Things to Be Careful About

Read values carefully from the scale (check each major division).
Keep consistent significant figures (typically 2–3 s.f. for gradient and intercept).
Ensure you are using the same $x$ unit as on the graph (cm vs m). If you plotted $x$ in cm, your gradient is in $\text{cm}^{-1}$ .

Techniques used

select two widely separated points on a best-fit linecalculate gradient as $\Delta y / \Delta x$read the y-intercept from where the line crosses the y-axis

(d)

It is suggested that the quantities $\theta$ and $x$ are related by the equation

\frac{1}{\tan \theta} = ax + b

where $a$ and $b$ are constants.

Using your answers in (c)(iii), determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______

$b$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{\tan\theta} = ax + b

Comparing with $y = mx + c$ for a graph of $y = 1/\tan\theta$ against $x$ :

a = \text{gradient},\quad b = y\text{-intercept}

Using (c)(iii) (example values):

a = 0.30\ \text{cm}^{-1}

b = 0.44

Answer

a \approx 0.30\ \text{cm}^{-1},\quad b \approx 0.44

Final answer

Example: a ≈ 0.30 cm⁻¹, b ≈ 0.44

Detailed explanation

Background Concept

A linear relationship

y = mx + c

means that when you plot $y$ against $x$ :

the gradient of the best-fit line is $m$
the $y$ -intercept is $c$

Units:

$y = 1/\tan\theta$ is dimensionless.
If $x$ is measured in $\text{cm}$ , then $m$ has units $\text{cm}^{-1}$ so that $mx$ is dimensionless.
The intercept $c$ has the same units as $y$ (dimensionless).

Understanding the Question

You are given the suggested equation:

\frac{1}{\tan\theta} = ax + b

and you have already found the gradient and y-intercept from the graph of $1/\tan\theta$ (y-axis) against $x$ (x-axis).

You are asked to determine $a$ and $b$ and include appropriate units.

Approach

Recognise that the graph is already in the form $y$ vs $x$ .
Identify $a$ with the gradient and $b$ with the y-intercept.
Attach correct units based on the units used on your axes.

Step-by-Step Reasoning

Let

y = \frac{1}{\tan\theta}

Then the given equation becomes:

y = ax + b

Comparing with $y = mx + c$ :

a = m\quad \text{and}\quad b = c

So:

$a$ is numerically equal to the gradient you found in (c)(iii).
$b$ is numerically equal to the y-intercept you found in (c)(iii).

Units:

Since $y$ is dimensionless and $x$ is in $\text{cm}$ , $a$ must be in $\text{cm}^{-1}$ .
$b$ is dimensionless (no unit).

Key Takeaways

The whole purpose of plotting the graph is to turn the relationship into $y = mx + c$ so constants can be read from gradient/intercept.
Always include units for constants that come from a gradient.

Common Mistakes

Swapping $a$ and $b$ .
Giving $b$ a unit when $y$ is dimensionless.
Giving $a$ the wrong unit because of changing $x$ units (e.g. using $\text{m}^{-1}$ when the graph used cm).

Things to Be Careful About

Use the units from your actual axes: if you plotted $x$ in $\text{m}$ , then $a$ would be in $\text{m}^{-1}$ .
Quote $a$ and $b$ to sensible significant figures consistent with the uncertainty in your graph readings (usually 2–3 s.f.).

Techniques used

match a straight-line equation to the form y = mx + cidentify constants as gradient and interceptdeduce units of constants from axis units

Q2MediumManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationPresentation of Data and Observations

In this experiment, you will investigate the motion of a conical pendulum.

(a)

Set the compasses to a radius of approximately $9\text{ cm}$ and then use them to draw a circle on the sheet of paper.
Mark the centre of the circle with a cross.
Measure and record the diameter $D$ of the circle.

$D$ = ______ $\text{cm}$

1M Similar

(b)

(i)

You are provided with a pendulum bob with a length of string attached.

Tie a knot in the string approximately $19\text{ cm}$ from the top of the bob.
Measure and record the distance $p$ from the knot to the centre of the bob.

$p$ = ______ $\text{cm}$

1M Similar

(ii)

Estimate the percentage uncertainty in your value of $p$ . Show your working.

percentage uncertainty = ______ %

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using a ruler with $1\ \text{mm}$ divisions, take absolute uncertainty $\Delta p \approx \pm 1\ \text{mm} = \pm 0.10\ \text{cm}$ .

\%\ \text{uncertainty} = \frac{\Delta p}{p} \times 100 = \frac{0.10}{20.0} \times 100 = 0.50\%

Answer

\text{percentage uncertainty in } p \approx 0.5\%

Final answer

0.5%

Detailed explanation

Background Concept

For a single ruler measurement, a common estimate of absolute uncertainty is about one smallest division (or sometimes half a division, depending on marking scheme/expectation). A ruler with $1\ \text{mm}$ resolution corresponds to $0.10\ \text{cm}$ . Percentage uncertainty is then:

\%\ \text{uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}}\times 100

Understanding the Question

You have measured $p$ and must estimate the percentage uncertainty in that value, showing working.

Approach

Decide a realistic absolute uncertainty for $p$ based on the instrument (ruler) and difficulty of judging the centre of the bob.
Convert that to a percentage using $\Delta p/p \times 100$ .

Step-by-Step Reasoning

If the ruler reads to $1\ \text{mm}$ , a reasonable absolute uncertainty is $\Delta p = \pm 1\ \text{mm} = \pm 0.10\ \text{cm}$ .
Using an example measurement $p = 20.0\ \text{cm}$ :

\frac{\Delta p}{p} = \frac{0.10}{20.0} = 0.005

\%\ \text{uncertainty} = 0.005 \times 100 = 0.50\%

So you would quote about $0.5\%$ .

Key Takeaways

Percentage uncertainty scales inversely with the measured value.
Always show the formula and substitution.

Common Mistakes

Using $\Delta p = 0.01\ \text{cm}$ (too small for a ruler) or using no justification.
Forgetting to multiply by $100$ .
Mixing units (e.g. $\Delta p$ in mm but $p$ in cm).

Things to Be Careful About

If you judge the centre of the bob by eye, the uncertainty might be larger than $1\ \text{mm}$ . In an exam, state a sensible value and then use it consistently.
Quote the final percentage to an appropriate number of significant figures (often 1 s.f. is fine for an uncertainty).

Techniques used

estimate absolute uncertainty from instrument resolutioncalculate percentage uncertainty using a ratio

(c)

(i)

Place the paper with the circle on the bench.
Holding the knot, suspend the bob approximately $5\text{ mm}$ above the cross at the centre of the circle, as shown in Fig. 2.1.

Fig. 2.1

Move the knot in small, slow circles so that the bob starts to move in a circle.
Adjust the movement of the knot until the bob moves just above the circle on the paper, as shown in Fig. 2.2.

Fig. 2.2

The period $T$ of the rotation of the bob is the time the bob takes to travel through one complete circle.

When this motion is steady, take measurements to determine $T$ .

$T$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Time $N$ revolutions ( $N \ge 10$ ) with a stopwatch and divide by $N$ ; repeat and average.

Example:

N = 10,\quad t = 8.50\ \text{s}

T = \frac{t}{N} = \frac{8.50}{10} = 0.850\ \text{s}

Answer

T = 0.850\ \text{s}\ \text{(example)}

Final answer

T = 0.850 s (example)

Detailed explanation

Background Concept

The period $T$ is the time for one complete cycle (here, one full circle of the bob). Stopwatch reaction time is a major source of random uncertainty, so timing one cycle is poor. Timing many cycles and dividing reduces the fractional (percentage) uncertainty.

Understanding the Question

You must obtain a value for the period of rotation $T$ when the bob moves steadily in a horizontal circle just above the drawn circle. The question awards marks for how you take measurements, not just a number.

Approach

Choose a number of revolutions $N$ (typically $10$ to $20$ ).
Start timing when the bob passes a fixed reference point on the circle.
Stop timing after exactly $N$ revolutions as it passes the same point.
Repeat at least once and average.

Step-by-Step Reasoning

Get the conical pendulum motion steady with the bob tracing the circle.
Pick a reference point (e.g. where the bob passes near a mark on the paper).
Measure time $t$ for $N = 10$ revolutions.
Calculate:

T = \frac{t}{N}

Example: if $t = 8.50\ \text{s}$ for $10$ revolutions, then $T = 0.850\ \text{s}$ .
5. Repeat (e.g. another $10$ revolutions) and average the two values of $T$ .

Key Takeaways

Timing many cycles reduces percentage uncertainty.
Use a consistent start/stop point to avoid counting errors.

Common Mistakes

Timing only one revolution (very large percentage uncertainty).
Starting and stopping at different points in the cycle.
Miscounting revolutions (especially if motion is not steady).

Things to Be Careful About

Ensure the bob stays close to the drawn circle radius while timing.
If the speed drifts, discard the run and re-adjust to steady motion before re-timing.
Record $T$ to a sensible precision (often 0.01 s or 0.001 s depending on how $t$ and $N$ are recorded).

Techniques used

time multiple revolutions and divide to find a periodrepeat timing measurements and average to reduce random erroruse consistent start/stop reference point when timing periodic motion

(ii)

The angle between the string and the vertical when the bob follows this circular path is $\Phi$ , where $\Phi$ is given by

\sin \Phi = \frac{D}{2p}

Calculate $\Phi$ .

$\Phi$ = ______ $^\circ$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\sin \Phi = \frac{D}{2p} = \frac{18.0}{2 \times 20.0} = 0.450

\Phi = \sin^{-1}(0.450) = 26.7^\circ

Answer

\Phi = 26.7^\circ

Final answer

26.7°

Detailed explanation

Background Concept

In a conical pendulum, the string makes an angle $\Phi$ with the vertical. If the bob moves in a horizontal circle of radius $r$ , then from right-triangle geometry:

\sin \Phi = \frac{r}{p}

Here the circle drawn has diameter $D$ , so $r = D/2$ , giving the provided relation:

\sin \Phi = \frac{D}{2p}

Understanding the Question

You must use your measured $D$ (circle diameter) and $p$ (knot to centre of bob) to calculate the angle $\Phi$ in degrees.

Approach

Compute the ratio $D/(2p)$ (make sure $D$ and $p$ are in the same units).
Take inverse sine to obtain $\Phi$ .
Quote the answer in degrees.

Step-by-Step Reasoning

Using example measured values $D = 18.0\ \text{cm}$ and $p = 20.0\ \text{cm}$ :

\sin \Phi = \frac{18.0}{2 \times 20.0} = \frac{18.0}{40.0} = 0.450

Then:

\Phi = \sin^{-1}(0.450) = 26.7^\circ

(Your number will differ slightly depending on your measured $D$ and $p$ .)

diagram-c-phi-geometry

Key Takeaways

The diameter gives the radius via $r = D/2$ .
Keep units consistent inside ratios.
Use $\sin^{-1}$ to undo a sine.

Common Mistakes

Using $D/p$ instead of $D/(2p)$ .
Mixing units (e.g. $D$ in cm and $p$ in m).
Giving the calculator result in radians instead of degrees.

Things to Be Careful About

Check that $D/(2p) \le 1$ ; if it is greater than 1, a measurement is inconsistent (likely $p$ too small or $D$ too large).
Quote $\Phi$ to a sensible precision (often 0.1° is adequate given measurement uncertainties).

Techniques used

substitute measured values into a trigonometric relationshiprearrange to find an angle using an inverse trigonometric functionconvert a computed angle into degrees

(d)

Tie a knot in the string approximately $13\text{ cm}$ from the top of the bob.
Using this knot, measure and record $p$ .

$p$ = ______ $\text{cm}$

Using this knot, repeat (c).

$T$ = ______

$\Phi$ = ______ $^\circ$

3M Similar

DifficultyMedium-Easy

Worked solution

Measure $p$ (nearest $0.1\ \text{cm}$ ), determine $T$ by timing $N$ revolutions and dividing, then calculate $\Phi$ using $\sin \Phi = D/(2p)$ .

Example set of results:

p = 14.0\ \text{cm}

N = 10,\quad t = 6.60\ \text{s} \Rightarrow T = \frac{6.60}{10} = 0.660\ \text{s}

\sin \Phi = \frac{18.0}{2 \times 14.0} = 0.643 \Rightarrow \Phi = \sin^{-1}(0.643) = 40.1^\circ

Final answer

Example: p = 14.0 cm, T = 0.660 s, \Phi = 40.1°

Detailed explanation

Background Concept

To test a proposed relationship experimentally, you need at least two sets of readings with different values of the relevant variable(s). Here, changing the knot position changes $p$ and therefore changes the geometry (and period) of the conical pendulum.

Understanding the Question

You now tie the knot closer to the bob (about $13\ \text{cm}$ from the top of the bob), so the distance $p$ from the knot to the centre of the bob is smaller. Using this new knot, you must repeat the measurements from part (c): obtain $T$ and then calculate $\Phi$ .

Approach

Measure $p$ carefully to the centre of the bob.
Generate steady conical motion so the bob travels above the same drawn circle.
Time many revolutions to find $T$ and repeat if possible.
Use the same diameter $D$ and compute $\Phi$ using $\sin\Phi = D/(2p)$ .

Step-by-Step Reasoning

Tie the knot at the new position and measure $p$ with a ruler (record to $0.1\ \text{cm}$ ).
Start the conical motion and adjust until the bob is just above the drawn circle.
Choose a value like $N = 10$ revolutions; measure total time $t$ and compute $T = t/N$ .
Use the measured $D$ and new $p$ :

\sin\Phi = \frac{D}{2p}

Then find $\Phi = \sin^{-1}(\,D/(2p)\,)$ .

Key Takeaways

Keep the radius of motion consistent with the drawn circle for both trials.
Timing multiple revolutions is essential for reasonable precision.

Common Mistakes

Forgetting to re-measure $p$ after tying the new knot.
Allowing the bob to trace a different radius than the drawn circle (changes $D/2$ effectively).
Timing too few revolutions.

Things to Be Careful About

Use the same value of $D$ (same circle) for both trials.
Ensure the motion is steady before timing; otherwise $T$ is not well-defined.

Techniques used

take a second set of readings with a changed independent variabletime multiple revolutions and divide to find a periodcalculate an angle from measured quantities using an inverse trigonometric function

(e)

It is suggested that the relationship between $T$ , $p$ and $\Phi$ is

T^2 = kp \cos \Phi

where $k$ is a constant.

(i)

Using your data, calculate two values of $k$ .

first value of $k$ = ______

second value of $k$ = ______

1M Similar

DifficultyMedium

Worked solution

Working

From

T^2 = kp\cos\Phi

k = \frac{T^2}{p\cos\Phi}

Using set 1 (example): $T = 0.850\ \text{s}$ , $p = 0.200\ \text{m}$ , $\Phi = 26.7^\circ$

k_1 = \frac{(0.850)^2}{0.200\cos 26.7^\circ} = 4.04\ \text{s}^2\ \text{m}^{-1}

Using set 2 (example): $T = 0.660\ \text{s}$ , $p = 0.140\ \text{m}$ , $\Phi = 40.1^\circ$

k_2 = \frac{(0.660)^2}{0.140\cos 40.1^\circ} = 4.07\ \text{s}^2\ \text{m}^{-1}

Answer

first value of $k = 4.04\ \text{s}^2\ \text{m}^{-1}$

second value of $k = 4.07\ \text{s}^2\ \text{m}^{-1}$

Final answer

k1 = 4.04 s^2 m^-1, k2 = 4.07 s^2 m^-1 (example)

Detailed explanation

Background Concept

If a relationship is proposed as

T^2 = kp\cos\Phi,

then the constant $k$ can be found by rearranging:

k = \frac{T^2}{p\cos\Phi}.

To check the relationship, you calculate $k$ for different sets of measurements. If the relationship is valid, $k$ should be approximately the same (within experimental uncertainty) each time.

Understanding the Question

You have two sets of experimental readings (from the two knot positions): each set gives you $T$ , $p$ , and $\Phi$ . You must calculate two values of $k$ using these data.

Approach

Rearrange the equation to make $k$ the subject.
Convert $p$ into SI units (metres) so the unit of $k$ is consistent.
Substitute each set of values into the formula and compute $k_1$ and $k_2$ .

Step-by-Step Reasoning

Start with:

T^2 = kp\cos\Phi

Divide both sides by $p\cos\Phi$ :

k = \frac{T^2}{p\cos\Phi}

Use consistent units. If $p$ is measured in cm, convert to m: $p(\text{m}) = p(\text{cm})/100$ .
Example calculations:

Set 1: $T = 0.850\ \text{s}$ , $p = 20.0\ \text{cm} = 0.200\ \text{m}$ , $\Phi = 26.7^\circ$ .

T^2 = (0.850)^2 = 0.7225\ \text{s}^2

p\cos\Phi = 0.200\cos 26.7^\circ = 0.200(0.894) = 0.179\ \text{m}

k_1 = \frac{0.7225}{0.179} = 4.04\ \text{s}^2\ \text{m}^{-1}

Set 2: $T = 0.660\ \text{s}$ , $p = 14.0\ \text{cm} = 0.140\ \text{m}$ , $\Phi = 40.1^\circ$ .

T^2 = (0.660)^2 = 0.4356\ \text{s}^2

p\cos\Phi = 0.140\cos 40.1^\circ = 0.140(0.765) = 0.107\ \text{m}

k_2 = \frac{0.4356}{0.107} = 4.07\ \text{s}^2\ \text{m}^{-1}

These are close, suggesting consistency.

Key Takeaways

Always rearrange first and then substitute.
Use SI units for derived constants unless the question clearly uses other units.
Consistent values of $k$ across trials support the proposed relationship.

Common Mistakes

Using $p$ in cm without conversion, giving a value of $k$ too small by a factor of $100$ .
Forgetting the $\cos\Phi$ term (using $k = T^2/p$ ).
Calculator in radians for cosine instead of degrees.

Things to Be Careful About

Use the same degree mode consistently for $\sin^{-1}$ and $\cos$ .
Ensure you square $T$ correctly and keep enough intermediate precision to avoid rounding error.

Techniques used

rearrange an equation to make the constant the subjectsubstitute experimental values consistently in SI unitscompute two values and compare for consistency

(ii)

Justify the number of significant figures that you have given for your values of $k$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$k$ is calculated from measured values of $T$ , $p$ and $\Phi$ . The overall uncertainty is about $15\%$ , so quoting $k$ to more than about 2 significant figures is not justified.

(e.g. $k = 4.0\ \text{s}^2\ \text{m}^{-1}$ to 2 s.f.)

Final answer

2 s.f. (because uncertainty ~15%)

Detailed explanation

Background Concept

Significant figures in a calculated quantity should reflect the uncertainty in the data used. As a rule of thumb:

If a result has an uncertainty of order $10\%$ – $20\%$ , then quoting more than about 2 significant figures is usually meaningless.
You should not claim precision your measurements do not support.

Understanding the Question

You calculated values of $k$ from $T^2/(p\cos\Phi)$ . The question asks you to justify the number of significant figures you used for $k$ .

Approach

Look at the measurement uncertainties/precision:

$T$ from stopwatch timing (often limited by reaction time and counting) typically dominates.
$p$ is a ruler measurement (often to $0.1\ \text{cm}$ ).
$\Phi$ is derived from $D$ and $p$ and so also has uncertainty.
Then relate this to the suggested overall uncertainty (later given as $15\%$ ).

Step-by-Step Reasoning

Because the experiment suggests the percentage uncertainty in $k$ is about $15\%$ , the last digit of $k$ is uncertain at about the first decimal place for a value near $4$ .

Quoting $k = 4.04$ implies an uncertainty of about $\pm 0.01$ (about $0.25\%$ ), which is far smaller than $15\%$ .
Quoting $k$ to 2 s.f., e.g. $k = 4.0\ \text{s}^2\ \text{m}^{-1}$ , matches a realistic experimental precision.

Key Takeaways

The quoted significant figures must match the uncertainty.
An uncertainty of $15\%$ supports only about 2 significant figures.

Common Mistakes

Quoting too many significant figures because the calculator shows them.
Justifying sig figs only from one measurement (e.g. $p$ ) while ignoring timing uncertainty in $T$ .

Things to Be Careful About

If you used $p$ in cm instead of m, the numerical value of $k$ changes, but the significant-figure argument (set by percentage uncertainty) is the same.
Keep your sig-fig choice consistent across both values of $k$ .

Techniques used

identify the limiting precision among measured quantitiesjustify significant figures based on experimental uncertainty

(f)

It is suggested that the percentage uncertainty in the values of $k$ is $15\%$ .

Using this uncertainty, explain whether your results support the relationship in (e).

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Using example values $k_1 = 4.04$ and $k_2 = 4.07$ :

\%\ \text{difference} = \frac{|k_1-k_2|}{\frac{1}{2}(k_1+k_2)}\times 100 = \frac{0.03}{4.055}\times 100 = 0.74\%

Since $0.74\% < 15\%$ , the two values agree within the stated uncertainty.

Answer

Yes. The values of $k$ are consistent within $15\%$ , so the results support the suggested relationship.

Final answer

Supports the relationship (agreement within 15%).

Detailed explanation

Background Concept

When testing a relationship that predicts a constant $k$ , you check whether values of $k$ from different trials are consistent. A common method is to compare the percentage difference between the values with the percentage uncertainty. If the difference is less than (or comparable to) the uncertainty, the results are consistent.

Understanding the Question

You are told the percentage uncertainty in $k$ is $15\%$ . You have two experimental values of $k$ from part (e)(i). You must state whether the results support the relationship.

Approach

Compute how far apart the two values of $k$ are (difference).
Convert that into a percentage difference (relative to a typical value such as the mean).
Compare with $15\%$ and conclude.

Step-by-Step Reasoning

Using example values $k_1$ and $k_2$ :

\%\ \text{difference} = \frac{|k_1-k_2|}{\text{mean}}\times 100

If $k_1 = 4.04$ and $k_2 = 4.07$ , the mean is $4.055$ and the difference is $0.03$ , so:

\%\ \text{difference} = \frac{0.03}{4.055}\times 100 \approx 0.74\%

Because $0.74\%$ is much less than $15\%$ , the two $k$ values agree within uncertainty. That supports the proposed relationship.

Key Takeaways

“Support” means agreement within uncertainty, not perfect equality.
Always compare differences to the stated uncertainty.

Common Mistakes

Comparing absolute difference (e.g. $0.03$ ) without converting to a percentage.
Using $\frac{|k_1-k_2|}{k_1}$ but not stating what you used (mean is usually clearer).
Concluding “proved” instead of “supports” (experiments provide evidence, not proof).

Things to Be Careful About

If your two values differ by less than about $15\%$ , say they support the relationship; if greater than $15\%$ , say they do not (within that uncertainty).
Use consistent rounding so you do not inflate the difference by premature rounding.

Techniques used

compare two values using a percentage differencejudge agreement using a stated percentage uncertaintystate a conclusion consistent with the comparison

(g)

(i)

Describe four sources of uncertainty or limitations of the procedure for this experiment.

For any uncertainties in measurement that you describe, you should state the quantity being measured and a reason for the uncertainty.

4M Similar

DifficultyMedium

Worked solution

Answer

Measuring $T$ : stopwatch reaction time and difficulty judging the exact instant the bob passes the reference point (uncertainty in timing).
Measuring $p$ : difficult to locate the centre of the bob accurately and possible parallax when reading the ruler (uncertainty in $p$ ).
Keeping the radius equal to $D/2$ : the bob may not move exactly above the drawn circle throughout; the radius can vary as the motion drifts (limitation affecting $D$ used for the actual path).
Motion not perfectly steady/circular: speed and angle can vary due to hand motion, air resistance and string twisting, so $T$ and $\Phi$ are not constant during a run (limitation in both $T$ and $\Phi$ ).

Final answer

See working (four uncertainties/limitations listed).

Detailed explanation

Background Concept

Uncertainties and limitations in a practical come from:

measurement resolution (instrument limits),
human judgement (reaction time, alignment, parallax),
the system not matching the ideal model (e.g. non-steady motion, changing radius),
uncontrolled variables (air resistance, pivot friction).

Marks are earned by being specific: name the quantity, and explain why it is uncertain.

Understanding the Question

You must describe four sources of uncertainty or procedural limitations in this conical pendulum experiment. For measurement uncertainties, you must specify what is being measured and why the measurement is uncertain.

Approach

Identify the main measured/derived quantities ( $D$ , $p$ , $T$ , $\Phi$ ) and then state realistic issues affecting each. Choose four distinct points and write each as “quantity + reason”.

Step-by-Step Reasoning

Here are four strong, distinct examples:

Timing the period $T$ : Using a stopwatch introduces reaction-time error. Also, it is hard to decide the exact moment the bob completes a full circle and passes the start point.
Measuring $p$ : The centre of the bob is not a sharp point, so locating it by eye adds uncertainty. Reading a ruler can also suffer from parallax if the ruler/eye is not perpendicular.
Radius of circular path: The method assumes the bob’s path radius is exactly $D/2$ , but during motion it may pass inside/outside the drawn circle. This means the effective $D$ for the motion is uncertain.
Non-steady conical motion: Because the knot is moved by hand, the bob may speed up/slow down and the cone angle may change. Air resistance and string twist can also alter the motion during timing.

(Any other sensible, clearly explained limitations such as pivot not fixed, string stretching, or bob not staying at constant height can also gain credit.)

Key Takeaways

Link each limitation to a specific measurement or to the validity of the model.
Explain the physical cause, not just “human error”.

Common Mistakes

Vague statements like “reaction time” without saying it affects timing $T$ .
Repeating the same idea in different words (e.g. “parallax” and “reading error” as separate points without a new quantity/reason).
Saying “air resistance” with no explanation of its effect (changes speed/steadiness).

Things to Be Careful About

Aim for four distinct points across different parts of the experiment (timing, lengths, maintaining radius, steadiness).
Avoid claiming the uncertainty is due only to instrument precision when the dominant effect is often human judgement (e.g. timing).

Techniques used

identify sources of random uncertainty and systematic errorlink each uncertainty to a specific measured quantitystate a physical reason for each limitation

(ii)

Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.

4M Similar

DifficultyMedium

Worked solution

Answer

Reduce timing uncertainty: time a larger number of revolutions (e.g. $20$ ) and repeat/average; or use video analysis/data logger to obtain period.
Improve $p$ measurement: mark the centre of the bob (or measure bob diameter with callipers and calculate centre) and use a set square to align the ruler to reduce parallax.
Keep radius constant: attach the knot to a fixed point (e.g. clamp stand with a low-friction ring/hook) and use a pointer/marker on the bob so you can keep the path exactly over the circle.
Improve steadiness of motion: use a smoother pivot (bearing/ring) and ensure the string is not twisting; allow motion to settle before timing and keep the bob at constant height above the paper.

Final answer

See working (four improvements listed).

Detailed explanation

Background Concept

Improvements are changes that specifically reduce random uncertainty, remove systematic error, or make the motion better match the theoretical model. The highest credit answers directly pair an improvement with a stated limitation.

Understanding the Question

You must describe four improvements to the experiment. You may change apparatus or procedure. The improvements should be practical and should clearly reduce uncertainty/limitations identified in part (g)(i).

Approach

Take each major limitation (timing, measuring $p$ , maintaining radius, steadiness) and propose an improvement that targets it.

Step-by-Step Reasoning

Four good improvements include:

Timing: Time $N=20$ (or more) revolutions and repeat at least twice, then average $T$ . This reduces percentage effect of reaction time. Alternatively, record motion with a phone camera and use frame-by-frame timing for one revolution or multiple revolutions.
Measuring $p$ : Measure the bob diameter with vernier callipers and mark/identify the centre accurately; measure from knot to that centre. Use a set square or align the ruler carefully to reduce parallax.
Maintaining a constant radius: Instead of holding the knot by hand, attach it to a fixed pivot point on a clamp stand (e.g. through a smooth ring). Use a visible marker on the bob and a clear circle target so you can keep the path directly above the drawn circle.
Steady conical motion: Use a low-friction pivot and ensure the string is free to rotate without twisting; start with small, controlled motion and allow it to settle before timing to minimise changes in $\Phi$ and speed.

Key Takeaways

Improvements should be specific and feasible.
Best answers: improvement clearly reduces a named uncertainty.

Common Mistakes

Saying “use better equipment” without naming what and how it helps.
Giving an improvement that does not address the described limitation.
Repeating the same improvement in different wording.

Things to Be Careful About

Avoid unrealistic suggestions (e.g. “eliminate air resistance”) unless you propose a plausible method.
If you suggest a data logger/video, state what quantity it measures (e.g. period from timestamps) and how it reduces human reaction error.

Techniques used

propose practical modifications that reduce stated uncertaintiesmatch each improvement to a specific limitationsuggest more precise instrumentation or procedures